In the diode circuit shown in the V1=10 sin 314.159 t V and VR= 5 V. Assume the diode to be ideal.
Right Answer is:
+5 V and −10V
SOLUTION
Given
Vin = 10 v
VR = 5 V
During the Positive half cycle, the diode is forward bias i.e Vin > Vr and the diode acts as a short circuit
Therefore the output voltage
Vout = Vin − Vr
Vout = 5 V (max Voltage)
During the negative half cycle, the diode is reversed biased and acts as an open circuit i.e Vin < Vr
Hence output voltage
Vout = Vin
Vout = −10 V (min voltage)